Seismic Loading and Analysis in Esteem 7

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Esteem Technical Manual

Seismic loading and analysis

References

1. Nawy, Edward G.Reinforced concrete: a fundamental approach / Edward G. Nawy. --4th ed., 2000Chap. 15

2. Farzad NaeimThe Seismic Design Handbook2nd Ed.,2001pp. 140ff

3. International Code CouncilInternational Building Code 2006 (IBC)

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Formulation of seismic parameters

The following table outlines the steps in obtaining the coefficient Cs (Seismic Load Factor)

and k(Seisimic Distribution Exponent) required for the seismic load application. The "Equivalent lateral force method" is used in Esteem 7 for earthquake loadings.

Steps Calculations Description Dependent Parameter Assumption

1. V = Cs.W V = seismic base shear

Cs = seismic response

coefficient or seismic loadfactor

W = The effective seismicweight fo the structure,inclusive or 20% liveloading.

2. Cs= SDS/(R/I) SDS = Design spectralresponse acceleration atshort period.

SDS=2/3.SMS

SMS=Fa.SS

SS=Mapped Spectralacceleration for shortperiods, eg. 0.075g whereg=9.81 N/kg.

Fa=Site coefficent from

Table 15.2aof Ref. (1).

5%dampeddesign

3. Condition 1:Cs=calculated in

(2) above cannot

exceed thefollowing,Cs= SD1/((R/I).T)

SD1 = Design spectral

response acceleration at1 second period.

SD1=2/3.SM1

SM1=Fv.S1

S1=Mapped Spectral

acceleration for 1 secondperiod, eg. 0.025g where

g=9.81 N/kg.

Fv=Site coefficent from

Table 15.2bof Ref. (1).

5%dampeddesign

4. Condition 2:Cs=calculated in

(2) above cannotbe taken less thanfollowing,Cs=0.044 SDS

In lieu of an analysis, anapproximate fundamentalperiod can be used,

Ta=CT.h(3/4)

h = Program willautomatically computethe total height ofbuilding.

R= Response modificationfactor (see Table 1617.6ref. (3))

I=Occupancy importancefactor

CT=Building PeriodCoefficient

5. Condition 3:For building and structures in seismic designcategories E or F and in buildings and

structures for which the 1-sec spectralresponse, S1 is equal to or greater than

0.6g,


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6. SeismicDistributionExponent, k

Based on building period,T calculated above.k=1 for T =2.5 secInterpolate linearly kvalue for T between 0.5and 2.5 sec.

Notes :1. Mapped Spectral accelerations, SS and S1(Spectral Response Acceleration) are to be

obtained from published regional geographic charts. In absence of such charts, engineersmust use good judgment accordingly. See Figure 1613.5ff of the "International BuildingCode (IBC) 2006" for US charts as example.

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Site Classification, Table 1613.5.2 ref. (3)

Notes :1. As far as Esteem 7 is concerned, site classification affects the calculation of Cs when

Site Class E & F are selected. It invites the consideration of Condition 3 as outline in thetopic above.

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Fa, Table 15.2a ref. (1)

Value of Site Coefficient Fa as a function of site class and mapped spectral response

acceleration at short periods (SS).

Site Class SS =< 0.25 SS =< 0.50 SS =< 0.75 SS =< 1.00 SS =< 1.25

A 0.8 0.8 0.8 0.8 0.8

B 1.0 1.0 1.0 1.0 1.0

C 1.2 1.2 1.1 1.0 1.0

D 1.6 1.4 1.2 1.1 1.0

E 2.5 1.7 1.2 0.9

F

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Fv, Table 15.2b ref. (1)

Value of Site Coefficient Fv as a function of site class and mapped spectral response

acceleration at 1.0 sec periods (S1).

Site Class S1 =< 0.1 S1 =< 0.2 S1 =< 0.3 S1 =< 0.4 S1 =< 0.5

A 0.8 0.8 0.8 0.8 0.8

B 1.0 1.0 1.0 1.0 1.0

C 1.7 1.6 1.5 1.4 1.3

D 2.4 2.0 1.8 1.6 1.5

E 3.5 3.2 2.8 2.4

F

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R, Table 1617.6 ref. (3)

extract of Table 1617.6 ref. (3) for Response modification coefficient R.

Basic Seismic-Force-Resisting System ResponseModificationCoefficient, R

System Limitations and Building Height Limitations (m)

by Seismic Deisgn Categorycas determined in IBCSection 1616.1

A&B C Dd Ee Ff

Bearing Wall System

Special reinforced concrete shear wall 5.5 NL NL 48.8 48.8 48.8

Ordinary reinforced concrete shear wall 4.5 NL NL NP NP NP

Detailed plain concrete shear walls 2.5 NL NL NP NP NP

Ordinary plain concrete shear walls 1.5 NL NP NP NP NP

Building Frame System NP NP NP

Ordinary reinforced concrete shear wall 5 NL NL NP NP NP

Detailed plain concrete shear walls 3 NL NL NP NP NP

Ordinary plain concrete shear walls 2 NL NP NP NP NP

Moment Resistant Frames

Special reinforced concrete moment frames 8 NL NL NL NL NL

Intermediate reinforced concrete momentframes

5 NL NL NP NP NP

Ordinary reinforced concrete momentframes

3 NLh NP NP NP NP

Dual System with Special MomentFrames

Special reinforced concrete shear walls 8 NL NL NL NL NL

Ordinary reinforced concrete shear walls 7 NL NL NP NP NP

Dual System with IntermediateMoment Frames

Special reinforced concrete shear walls 6 NL NL 48.8 30.5 30.5

Ordinary reinforced concrete shear walls 5.5 NL NL NP NP NP

Shear wall-frame Interactive system withordinary reinforced concrete moementframes and ordinary reinforced concreteshear walls

5.5 NL NP NP NP NP

c NL = not limited and NP = not permittedd limited to buildings with a height of 73.2m or less.e limited to buildings with a height of 48.8m or less.fOrdinary moment frame is permitted to be used in lieu of Intermediate moment frame in seismic design categoriesB, and C.h Ordinary moment frames of reinforced concrete are not permitted as a part of the seismic-force-resisting systemin seismic design category B structures fonded on Site Class E or F soils

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I, Table 15.5 ref. (1)

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Examples

Example to Ref. 1 (Nawy) page 693

A building of 5 storey of 2.9m height per floor (total height = 14.5m).

Building category II Site Class B

Seismic group II (use in spectral response method)

Response modification method, R = 3.0Occupancy importance factor, I = 1.25

From ground motion maps, S1=0.42, Ss=0.85 (Site B with 5% damping)

Adjusted spectral response accelerations for site class effects:

o for S1=0.42, Fv=1.0

o for SS=0.85, Fa=1.0o SMS=Fa.SS=1.0 x 0.85=0.85

o SM1=Fv.S1=1.0 x 0.42=0.42

o SDS=2/3 . SMS=2/3 x 0.85 = 0.567

o SD1=2/3 . SM1=2/3 x 0.42 = 0.278

CS=SDS/(R/I)=0.567/(3/1.25)=0.236 but cannot exceed

CS=SD1/(R/I) . T

; Ta= CTh(3/4)= 0.085 {for moment resisting frames} . 14.5(3/4)= 0.63sec

Hence, CS=SD1/(R/I) . T = 0.278 /((3/1.25) x 0.63) = 0.184

k= 1+ (0.63-0.5)/(2.5-0.5)=1.065

Example 2

A building of 5 storey of 3m height per floor (total height = 15m).

Building category II

Site Class B

Seismic group II (use in spectral response method)

Response modification method, R = 3.0Occupancy importance factor, I = 1.25

From ground motion maps, S1=0.03, Ss=0.075 (Site B with 5% damping)

Adjusted spectral response accelerations for site class effects:

o for S1=0.03, Fv=0.8

o for SS=0.075, Fa=0.8

o SMS=Fa.SS=0.8 x 0.075=0.06

o SM1=Fv.S1=0.8 x 0.03=0.024

o SDS=2/3 . SMS=2/3 x 0.06 = 0.04

o SD1=2/3 . SM1=2/3 x 0.024 = 0.016

CS=SDS/(R/I)=0.04/(3/1.25)=0.0167 but cannot exceed

CS=SD1/(R/I) . T


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; Ta= CTh(3/4)= 0.085 {for moment resisting frames} . 15(3/4)= 0.648 sec

Hence, CS=SD1/(R/I) . T = 0.016 /((3/1.25) x 0.648) = 0.01

k= 1+ (0.648-0.5)/(2.5-0.5)=1.074

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CT, Building Period Coefficient, ref. (1)

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Vertical Distribution of Forces

The lateral force Fx induced at any level can be determined from the following expressions

illustrated by Nawy.

Cvx = vertical distribution factor

V = total design lateral force or shear at the base of the building.Wiand Wx= the portion of the total gravity load of the building, W, located or assigned to

Level i or xhiand hx= the height from the base to level i or x

k = seismic distribution exponent

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Seismic Loading and Analysis in Esteem 7